The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + ........$, will be

If $a, b, c, d$ and $p$ are different real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right)\, \leq \,0,$ then show that $a, b, c$ and $d$ are in $G.P.$

If $n$ geometric means be inserted between $a$ and $b$ then the ${n^{th}}$ geometric mean will be

In an increasing geometric progression ol positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is :-

If $a$,$b$,$c \in {R^ + }$ are such that $2a$,$b$ and $4c$ are in $A$.$P$ and $c$,$a$ and $b$ are in $G$.$P$., then

If $y = x - {x^2} + {x^3} - {x^4} + ......\infty $, then value of $x$ will be